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Is this true in general? (Without assuming existence of variance etc.). Only thing we know is that the r.v.s are positive with a non-zero probability and that they are integer valued, integrable with mean 0.

Only thing I could come up with was that $S_n/n$ goes to 0 (where $S_n$ is the partial sum). But this can happen even if $S_n$ was always positive.

My intuition is that this should be true because it happens with Simple Random Walks (which have a unit step size), and a larger step size should only increase the probability of it becoming negative if it was positive before. I am not sure how to formalize this. Can someone please help?

$S_n=X_1+dots+X_n$. This proof will work without the assumption that $X_i$ is integer valued. Let
$$
beginarraycc
alpha=inf n>0:S_n>0 &&beta=inf n>0:S_n<0\
alpha'=inf n>0:S_nge 0 &&beta'=inf n>0:S_nle 0
endarray
$$

Using Wald's equation, you can show that
$$
E[alpha]=E[beta]=infty.
$$
Otherwise, you would have $E[S_alpha]=E[X_1]E[alpha]=0cdot E[alpha]=0$, contradicting $S_alpha>0$.

Next is the real tricky part. For each $nge 0$, define $I_n$ to be the index $iin0,1,dots,n$ for which $S_i$ is minimized, with ties going to the latest such index. I claim that

$$
P(I_n=m)=P(alpha>m)P(beta'>n-m).
$$

To see this, note that $I_n=m$ is determined by the first $n$ steps $X_1,dots,X_n$ of the process, while $alpha>m$ is determined by the first $m$ steps and $beta'>n-m$ by the first $n-m$ steps. Furthermore, $I_n=m$ occurs for $(X_1,dots,X_n)$ if and only if $alpha>m$ occurs for $(X_m,X_m-1,dots,X_1)$ (note the reversal!) and $beta'>n-m$ occurs for $(X_m+1,X_m+2,dots,X_n)$. $square$

The payoff to that tricky Lemma is we can show

$alpha'$ and $beta'$ are almost surely finite.

We start with $$
1=sum_m=0^n P(I_n=m)=sum_m=0^n P(alpha>m)P(beta'>n-m)
$$
Now, let $ntoinfty$. Each summand $ P(alpha>m)P(beta'>n-m)$ converges to $P(alpha>m)P(beta'=infty)$, so we get
$$
E[alpha]=sum_m=0^infty P(alpha>m)=frac1P(beta'=infty)
$$
We already proved $E[alpha]=infty$, so this shows that $beta'$ is almost surely finite. The same goes for $alpha'$.

Finally,

$S_n$ is positive and negative infinitely often.

Just like $alpha'$ is the first time after $0$ the process is nonnegative, we define $alpha'(k)$ inductively to be the first time after $alpha'(k-1)$ that $S_nge S_alpha'(k-1)$. Now, the sequence of steps
$$
X_alpha'+1,X_alpha'(2)+1,X_alpha'(3)+1,dots
$$
are iid distributed like $X_1$. Assuming $X_1$ is nontrivial, these have a nonzero probability of being positive, so with probability $1$, infinitely many of them are positive. If $X_alpha'(k)+1$ is positive, then $S_alpha'(k)+1=S_alpha'(k)+X_alpha'(k)+1>0$. Therefore, there are infinitely many times of the form $S_alpha'(k)+1$ which are greater than $0$. The same goes for times where $S_n<0$.